我有一个具有不同值的一维numpy数组(arr0)。 我想创建一个新的元素数组,其中每个元素是一个元素与其最接近的元素的一对(索引和/或值),考虑到这对元素的差(距离)的绝对值小于一组阈值。
在每个步骤(耦合)中,我想删除已经耦合的元素。
arr0 = [40, 55, 190, 80, 175, 187] #My original 1D array
threshold = 20 #Returns elements if "abs(el_1 - el_2)<threshold"
#For each couple found, the code should remove the couple from the array and then go on with the next couple
result_indexes = [[0, 1], [2, 5]]
result_value = [[40, 55], [190, 187]]
答案 0 :(得分:0)
您可以使用itertools软件包中的combinations
来想象这样的事情。我还使用sklearn.metrics.pairwise_distances
计算所有成对距离:
from itertools import combinations
from sklearn.metrics import pairwise_distances
# Get all pairwise distances
distances = pairwise_distances(np.array(arr0).reshape(-1,1),metric='l1')
# Sort the neighbors by distance for each element
neighbors_matrix = np.argsort(distances,axis=1)
result_indexes = []
result_values = []
used_indexes = set()
for i, neighbors in enumerate(neighbors_matrix):
# Skip already used indexes
if i in used_indexes:
continue
# Remaining neighbors
remaining = [ n for n in neighbors if n not in used_indexes and n != i]
# The closest non used neighbor is in remaining[0] is not empty
if len(remaining) == 0:
continue
if distances[i,remaining[0]] < threshold:
result_indexes.append((i,remaining[0]))
result_values.append((arr0[i],arr0[remaining[0]]))
used_indexes = used_indexes.union({i,remaining[0]})
在您的示例中,它产生:
>> result_indexes
[(0, 1), (2, 4)]
>> result_values
[(40, 55), (190, 175)]
答案 1 :(得分:0)
arr0s = sorted(arr0)
n = len(arr0)
z = []
x = 0
while x<n-2:
if arr0s[x+1]-arr0s[x] < 20:
if arr0s[x+1]-arr0s[x] < arr0s[x+2]-arr0s[x+1]:
z.append([arr0s[x], arr0s[x+1]])
x+=2
else:
z.append([arr0s[x+1], arr0s[x+2]])
x+=3
else:
x+=1
result_indexes = [[arr0.index(i[0]), arr0.index(i[1])] for i in z]
for i, j in enumerate(result_indexes):
if j[0]>j[1]:
result_indexes[i] = [j[1], j[0]]
result_value = [[arr0[i[0]], arr0[i[1]]] for i in result_indexes]
print(result_indexes)
#[[0, 1], [2, 5]]
print(result_value)
#[[40, 55], [190, 187]]